Respuesta :
Answer:
A)Mass of gallium plated out is 0.3440 grams
B) For 0.67 hours current of 5.79 A must to be applied to plate out 8.70 g of tin.
Explanation:
To calculate the total charge, we use the equation:
[tex]C=I\times t[/tex]
where,
C = Charge
I = Current in time t (seconds)
To calculate the moles of electrons, we use the equation:
[tex]\text{Moles of electrons}=\frac{C}{F}[/tex]
where,
F = Faraday's constant = 96500
A) The equation for the deposition of Ga(s) from Ga(III) solution follows:
[tex]Ga^{3+}(aq.)+3e^-\rightarrow Ga(s)[/tex]
I = 0.790 A, t = 30.0 min = 1800 seconds
[tex]C=I\times t[/tex]
[tex]C=0.790 A\times 1800 s=1422 C[/tex]
Moles of electron transferred:
[tex]=\frac{1422 C}{96500 F}=0.01474 mol[/tex]
Now, to calculate the moles of gallium, we use the equation:
[tex]\text{Moles of Gallium}=\frac{\text{Moles of electrons}}{n}[/tex]
n = number of electrons transferred = 3
[tex]\text{Moles of Gallium}=\frac{0.01474 mol}{3}=0.004913 mol[/tex]
Mass of 0.004913 moles of gallium = 0.004913 mol × 70 g/mol=0.3440 g
B) The equation for the deposition of Sn(s) from Sn(II) solution follows:
[tex]Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)[/tex]
Moles of tin = [tex]\frac{8.70 g}{119 g/mol}=0.07311 mol[/tex]
n = number of electrons transferred = 2
[tex]\text{Moles of tin}=\frac{\text{Moles of electrons}}{n}[/tex]
Moles of electron = [tex]n\times \text{Moles of tin}[/tex]
[tex]=2\times 0.07311 mol=0.14622 mol[/tex]
Charge transferred during time t :
[tex]\text{Moles of electrons}=\frac{C}{F}[/tex]
[tex]C=96500 F\times 0.14622 mol=14,110.23 C[/tex]
Current applied for t time = I = 5.79 A
[tex]t=\frac{C}{I}=\frac{14,110.23 C}{5.79 A}=2,437 s=0.67 hrs[/tex]
For 0.67 hours current of 5.79 A must to be applied to plate out 8.70 g of tin.
